Practical E - Exercises

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The following packages are required for this practical:

library(mice)
library(dplyr)
library(magrittr)
library(stringr)
library(ggplot2)

Exercise 0

Function plot() is the core plotting function in R. Find out more about plot(): Try both the help in the help-pane and ?plot in the console. Look at the examples by running example(plot).

Solution

The help tells you all about a functions arguments (the input you can specify), as well as the element the function returns to the Global Environment. There are strict rules for publishing packages in R. For your packages to appear on the Comprehensive R Archive Network (CRAN), a rigorous series of checks have to be passed. As a result, all user-level components (functions, datasets, elements) that are published, have an acompanying documentation that elaborates how the function should be used, what can be expected, or what type of information a data set contains. Help files often contain example code that can be run to demonstrate the workings.

?plot

## Help on topic 'plot' was found in the following packages:
## 
##   Package               Library
##   graphics              /Library/Frameworks/R.framework/Versions/4.3-x86_64/Resources/library
##   base                  /Library/Frameworks/R.framework/Resources/library
## 
## 
## Using the first match ...

example(plot)

## 
## plot> Speed <- cars$speed
## 
## plot> Distance <- cars$dist
## 
## plot> plot(Speed, Distance, panel.first = grid(8, 8),
## plot+      pch = 0, cex = 1.2, col = "blue")

## 
## plot> plot(Speed, Distance,
## plot+      panel.first = lines(stats::lowess(Speed, Distance), lty = "dashed"),
## plot+      pch = 0, cex = 1.2, col = "blue")

## 
## plot> ## Show the different plot types
## plot> x <- 0:12
## 
## plot> y <- sin(pi/5 * x)
## 
## plot> op <- par(mfrow = c(3,3), mar = .1+ c(2,2,3,1))
## 
## plot> for (tp in c("p","l","b",  "c","o","h",  "s","S","n")) {
## plot+    plot(y ~ x, type = tp, main = paste0("plot(*, type = \"", tp, "\")"))
## plot+    if(tp == "S") {
## plot+       lines(x, y, type = "s", col = "red", lty = 2)
## plot+       mtext("lines(*, type = \"s\", ...)", col = "red", cex = 0.8)
## plot+    }
## plot+ }

## 
## plot> par(op)
## 
## plot> ##--- Log-Log Plot  with  custom axes
## plot> lx <- seq(1, 5, length.out = 41)
## 
## plot> yl <- expression(e^{-frac(1,2) * {log[10](x)}^2})
## 
## plot> y <- exp(-.5*lx^2)
## 
## plot> op <- par(mfrow = c(2,1), mar = par("mar")-c(1,0,2,0), mgp = c(2, .7, 0))
## 
## plot> plot(10^lx, y, log = "xy", type = "l", col = "purple",
## plot+      main = "Log-Log plot", ylab = yl, xlab = "x")

## 
## plot> plot(10^lx, y, log = "xy", type = "o", pch = ".", col = "forestgreen",
## plot+      main = "Log-Log plot with custom axes", ylab = yl, xlab = "x",
## plot+      axes = FALSE, frame.plot = TRUE)

## 
## plot> my.at <- 10^(1:5)
## 
## plot> axis(1, at = my.at, labels = formatC(my.at, format = "fg"))
## 
## plot> e.y <- -5:-1 ; at.y <- 10^e.y
## 
## plot> axis(2, at = at.y, col.axis = "red", las = 1,
## plot+      labels = as.expression(lapply(e.y, function(E) bquote(10^.(E)))))
## 
## plot> par(op)

There are many more functions that can plot specific types of plots. For example, function hist() plots histograms, but falls back on the basic plot() function. Packages lattice and ggplot2 are excellent packages to use for complex plots. Pretty much any type of plot can be made in R. A good reference for packages lattice that provides all R-code can be found at http://lmdvr.r-forge.r-project.org/figures/figures.html. Alternatively, all ggplot2 documentation can be found at http://docs.ggplot2.org/current/

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Exercise 1A

Create a scatter plot between age and bmi in the mice::boys data set. Use the base R plotting device.

Solution

With the standard plotting device in R:

plot( boys$bmi ~ boys$age )

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Exercise 1B

Create a scatter plot between age and bmi in the mice::boys data set. Use ggplot.

Hint

The point geom (geom_points()) is used to create scatterplots.

Solution

p <- ggplot( data = boys, aes(age, bmi))
p + geom_point()

## Warning: Removed 21 rows containing missing values (`geom_point()`).

Package ggplot2 offers far greater flexibility in data visualization than the standard plotting devices in R. However, it has its own language, which allows you to easily expand graphs with additional commands. To make these expansions or layers clearly visible, it is advisable to use the plotting language conventions. For example,

  ggplot( data = mice::boys, aes(age, bmi)) +
  geom_point()

would yield the same plot as

ggplot(mice::boys, aes(age, bmi)) + geom_point()

but the latter style may be less informative, especially if more customization takes place and if you share your code with others.

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Exercise 2A

Create a histogram for age in the boys data set. Use the base R plotting device.

Solution

With the standard plotting device in R:

hist(boys$age, breaks = 50)

The breaks = 50 overrides the default breaks between the bars. By default the plot would be

hist(boys$age)

The title and axis label need to be fixed:

hist(boys$age, breaks = 50, xlab = "Age", main = "Histogram")

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Exercise 2B

Create a histogram for age in the boys data set. Use ggplot.

Hint

The geom geom_histogram() is used to create histograms.

Solution

ggplot( data = boys ) + 
  geom_histogram(aes(age), binwidth = .4)

Please note that the plots from geom_histogram() and hist use different calculations for the bars (bins) and hence may look slightly different.

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Exercise 2.1A

Create a bar chart for reg in the boys data set. Use the base R plotting device.

Solution

With a standard plotting device in R:

boys %$%
  table(reg) %>%
  barplot()

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Exercise 2.1B

Create a bar chart for reg in the boys data set. Use ggplot.

Hint

The geom geom_bar() is used to create bar charts.

Solution

ggplot( data = boys ) + 
  geom_bar(aes(reg))

Note that geom_bar by default plots the NA's, while barplot() omits the NA's without warning. If we would not like to plot the NAs, then a simple filter() (see exercise 2) on the boys data is efficient.

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Exercise 2.2A

Create a box plot for hgt with different boxes for reg in the boys data set. Use the base R plotting device.

Solution

With a standard plotting device in R:

boys %$%
  boxplot(hgt ~ reg)

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Exercise 2.2B

Create a box plot for hgt with different boxes for reg in the boys data set. Use ggplot.

Hint

The geom geom_boxplot() is used to create box plots.

Solution

ggplot( data = boys, aes(reg, hgt)) +
  geom_boxplot()

## Warning: Removed 20 rows containing non-finite values (`stat_boxplot()`).

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Exercise 2.3A

Create a density plot for age with different curves for boys from the city and boys from rural areas (!city). Use the base R plotting device.

Solution

With a standard plotting device in R:

d1 <- boys %>%
  subset(reg == "city") %$%
  density(age)
d2 <- boys %>%
  subset(reg != "city") %$%
  density(age)

plot(d1, col = "red", ylim = c(0, .08))
lines(d2, col = "blue")

The above plot can also be generated without pipes, but results in an ugly main title. You may edit the title via the main argument in the plot() function.

plot(density(boys$age[!is.na(boys$reg) & boys$reg == "city"]),
     col = "red",
     ylim = c(0, .08))
lines(density(boys$age[!is.na(boys$reg) & boys$reg != "city"]),
      col = "blue")

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Exercise 2.3B

Create a density plot for age with different curves for boys from the city and boys from rural areas (!city). Use ggplot.

Hint 1

Create a new variable that indicates whether the area is urban or rural. Filter out missing values.

Hint 2

The geom geom_density() is used to create density plots.

Solution

With ggplot2:

boys %>%
  mutate(area = ifelse(reg == "city", "city", "rural")) %>%
  filter(!is.na(area)) %>%
  ggplot(aes(age, fill = area)) +
  geom_density(alpha = .3) # some transparency

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Exercise 3

Now recreate the plot from Exercise 1B with the following specifications:

• If bmi < 18.5 use color = "light blue"
• If bmi > 18.5 & bmi < 25 use color = "light green"
• If bmi > 25 & bmi < 30 use color = "orange"
• If bmi > 30 use color = "red"

Hint 1

It may help to expand the data set with a new variable.

Hint 2

It may be easier to create a new variable that creates the specified categories. We can use mutate() and the cut() function to do this quickly

Solution

boys2 <- 
  boys %>%
  mutate(class = cut(bmi, c(0, 18.5, 25, 30, Inf),
                    labels = c("underweight",
                               "healthy",
                               "overweight",
                               "obese")))

by specifying the boundaries of the intervals. In this case we obtain 4 intervals: 0-18.5, 18.5-25, 25-30 and 30-Inf. We can now call ggplot

ggplot(data = boys2) +
  geom_point(aes(age, bmi, col = class))

## Warning: Removed 21 rows containing missing values (`geom_point()`).

Although the different classifications have different colours, the colours are not conform the specifications of this exercise. We can manually override this:

ggplot(data = boys2 ) +
  geom_point(aes(age, bmi, col = class)) +
  scale_color_manual(values = c("light blue", "light green", "orange", "red"))

## Warning: Removed 21 rows containing missing values (`geom_point()`).

Because there are missing values, ggplot2 displays a warning message. If we would like to not consider the missing values when plotting, we can simply exclude the NAs by using a filter():

  ggplot( data = boys2[ !is.na( boys2$class ), ] ) +
  geom_point(aes(age, bmi, col = class)) +
  scale_color_manual(values = c("light blue", "light green", "orange", "red"))

Specifying a filter on the feature class is sufficient: age has no missings and the missings in class directly correspond to missing values on bmi. Filtering on bmi would therefore yield an identical plot.

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Exercise 4

Create a diverging bar chart for hgt in the boys data set, that displays for every age year that year's mean height in deviations from the overall average hgt. Use ggplot.

Hint 1

First create 2 new variabels. One for the height deviations and one for the (22?) age categories.

boys %>%
  mutate(Hgt = hgt - mean(hgt, na.rm = TRUE),
         Age = cut(age, 0:22, labels = 0:21)) 

Hint 2

Find the mean height deviations in each age group

group_by(Age)
  summarize(Hgt = mean(Hgt, na.rm = TRUE))

Hint 3

Define a new variable that indikates wether the height deviations is below or above average. You can apply cut().

mutate(Diff = cut(Hgt, c(-Inf, 0, Inf),
                  labels = c("Below Average", "Above Average")))

Solution

boys %>%
  mutate(Hgt = hgt - mean(hgt, na.rm = TRUE),
         Age = cut(age, 0:22, labels = 0:21)) %>%
  group_by(Age) %>%
  summarize(Hgt = mean(Hgt, na.rm = TRUE)) %>% 
  mutate(Diff = cut(Hgt, c(-Inf, 0, Inf),
                    labels = c("Below Average", "Above Average"))) %>%
  ggplot(aes(x = Age, y = Hgt, fill = Diff)) + 
  geom_bar(stat = "identity") +
  coord_flip()

We can clearly see that the average height in the group is reached just before age 7.

The group_by() and summarize() function are advanced dplyr functions used to return the mean() of deviation Hgt for every group in Age. For example, if we would like the mean and sd of height hgt for every region reg in the boys data, we could call:

boys %>%
  group_by(reg) %>% 
  summarize(mean_hgt = mean(hgt, na.rm = TRUE), 
            sd_hgt   = sd(hgt, na.rm = TRUE))

## # A tibble: 6 × 3
##   reg   mean_hgt sd_hgt
##   <fct>    <dbl>  <dbl>
## 1 north    152.    43.8
## 2 east     134.    43.2
## 3 west     130.    48.0
## 4 south    128.    46.3
## 5 city     126.    46.9
## 6 <NA>      73.0   29.3

The na.rm argument ensures that the mean and sd of only the observed values in each category are used.

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Exercise 5

Read in the sf package, and open the shapefiles on the Danish municipalities from the course homepage. Plot the REGIONNAVN variable to see the Danish regions. Plot the municipal-level population. Use ggplot.

Solution

library(sf)

## Linking to GEOS 3.11.0, GDAL 3.5.3, PROJ 9.1.0; sf_use_s2() is TRUE

denmark <- st_read("DK_map.shp")

## Reading layer `DK_map' from data source 
##   `/Users/mikkelmollerup/Dropbox/Work/RWORK/DSTStuff/Vietnam/RVietnam/Contents/Part_E/DK_map.shp' 
##   using driver `ESRI Shapefile'
## Simple feature collection with 306 features and 6 fields
## Geometry type: POLYGON
## Dimension:     XY
## Bounding box:  xmin: 441524.8 ymin: 6049785 xmax: 892800.8 ymax: 6402308
## CRS:           NA

class(denmark)

## [1] "sf"         "data.frame"

ggplot( data = denmark ) + geom_sf()

Say we want to colour the maps by the administrative regions coded in the REGIONNAVN variable. Also, we could move the legend inside the plot and use a more colorblind-friendly color scale:

denmark %>% ggplot( aes( fill = REGIONNAVN  ) ) + geom_sf() +
  theme(legend.position = c(0.8,0.7)) +
  scale_fill_brewer(palette = "Set2")

We plot the municipal-level population

denmark %>% mutate( population = population/1000) %>% 
  ggplot(aes( fill = population)) +
  geom_sf() + 
  scale_fill_viridis_c() + # The viridis color scale gives more visual nuance
  labs( fill = "Population,\nthousands")

But perhaps population per square kilometer might be more informative than just population

denmark$area <- st_area(denmark)/(1000^2)

denmark %>% 
  group_by(KOMKODE) %>% 
  summarise( total.area = sum( as.numeric( area ) ), 
             population = first( population )) %>% 
  mutate( pop.area = population/total.area) %>% 
  ggplot( aes( fill = pop.area)) + 
  geom_sf()+ 
  scale_fill_viridis_c() + # The viridis color scale gives more visual nuance
  labs( fill = "Population per \nsquare kilometer")

Our sf object contains more than one feature for some of the municipalities. The population number given is for the total municipality, so we need to compute the total area for each municipality.

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End of Practical

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Useful References

The ggplot2 reference page

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